The power dissipated is simply V^2/R
where V = 120 volts RMS
and R = 60 Ω
Answer:
JC⁻¹
Explanation:
= mass of water added to calorimeter = 94.8 g
= initial temperature of the water added = 60.4 C
= specific heat of water = 4.184 Jg⁻¹C⁻¹
= mass of water available to calorimeter = 94.8 g
= initial temperature of the water in calorimeter = 22.3 C
= final equilibrium temperature = 35 C
= Heat gained by calorimeter
Using conservation of heat
Heat gained by calorimeter = Heat lost by water added - heat gained by water in calorimeter
J
= Change in temperature of calorimeter
Change in temperature of calorimeter is given as
C
Heat capacity of calorimeter is given as
JC⁻¹
From Hooke's law the spring constant k is given by F = ke; then k = F/e. But F = Ma = Mg = 10 * 9.8 = 98N where g = 9.8ms^(-2) Extension e = 2cm = 0.02m. So k = 98/0.02 = 4900 N/m.
F = force by which Jon pushes the lawnmower = 50.0 N
θ = Angle of the force applied with the horizontal = 50 degrees
d = displacement of the lawnmower due to force applied = 90.0
W = work done on the lawnmower = ?
work done on the lawnmower is given as
W = F d Cosθ
inserting the values
W = (50.0) (90.0) Cos50
W = (4500) Cos50
W = 2892.54 J
t = time taken by Jon to mow the lawnmower = 10.0 minutes = 10.0 x 60 sec = 600 sec (Since 1 min = 60 sec)
Power is given as
P = W/t
inserting the values in the above formula
P = 2892.54/600
P = 4.82 Watt
